Projectile Motion without Initial velocity

[Disoo]

Homework Statement

A stone is fired from a slingshot at an angle of 65° from the horizontal. The stone strikes the ground 8.0s later at an altitude 30m lower than the height at which it was released.
At what initial velocity was the stone released?

Homework Equations

dy = viyt + 1/2gt2
Viy = Vi sin θ

The Attempt at a Solution

dy = viyt + 1/2gt2
dy = Vi sin θt + 1/2gt2
dy + 30m = Vi sin (65)(8.0s) + 1/2(+9.91m/s2)(8.0s)2
dy =7.25 Vi + 283.92 m

I'm stuck after this stage. Currently I don't see a clear solution. How could I approach the problem?

 

[rl.bhat]

Hi Disoo, welcome to PF.
In your equation you have not used the proper sign convention.
For the projectile motion the equation can be written as
y = yo + vi*sinθ*t - 1/2*g*t^2.
Take y = 0 at ground level.